PAT 1015 Reversible Primes (判断素数)

A reversible prime in any number system is a prime whose "reverse" in that number system is also a prime. For example in the decimal system 73 is a reversible prime because its reverse 37 is also a prime.

Now given any two positive integers N (<105) and D (1<D≤10), you are supposed to tell if N is a reversible prime with radix D.

Input Specification:

The input file consists of several test cases. Each case occupies a line which contains two integers N and D. The input is finished by a negative N.

Output Specification:

For each test case, print in one line Yes if N is a reversible prime with radix D, or No if not.

Sample Input:

73 10
23 2
23 10
-2

Sample Output:

Yes
Yes
No

思路

如果一个数为素数，且其在D进制下的反转过后的数字在10进制i下为素数，就输出Yes，否则输出No。

注意1为合数，1不是素数。会有一组数据关于这个的。

代码

#include <stdio.h>
#include <string>
#include <stdlib.h>
#include <iostream>
#include <vector>
#include <string.h>
#include <algorithm>
#include <cmath>
#include <map>
#include <limits.h> 
using namespace std;
bool flag[500000];
void init(){
	flag[1] = 1;
	for(int i = 2; i * i < 500000; i++){
		if(!flag[i]){
			for(int j = i * i; j < 500000; j += i){
				flag[j] = 1;
			}
		}
	}
}

int getNum(int N, int D){
	string t = "";
	while(N != 0){
		t = t + char(N % D);
		N /= D;
	}
	long long sum = 0;
	int d = 1;
	for(int i = t.length() - 1; i >= 0; i--){
		sum += t[i] * d;
		d *= D;
	}
	return sum;
}

int main() {
	int N, D;
	init();
	while(cin >> N){
		if(N < 0)	return 0;
		cin >> D;
		getNum(N, D); 
		if(!flag[N] && !flag[getNum(N, D)])	cout << "Yes" << endl;
		else cout << "No" << endl;
	}
	return 0; 
}